What's wrong with defining 0^0=1? As anyone familiar with the Binomial Theorem knows, it is a very convenient notion and avoids having to consider as a separate case expressions such as (2+0)^0 (even if it would be trivial to do so). (See http://en.wikipedia.org/wiki/Binomial_theorem) It is even possible, by various means (e.g. my Theorem 1 here), to formally construct or prove the existence of a binary function ^ on N such that

1. x^0 = 1 2. x^(y+1) = x^y * x

and to derive from this definition the usual Laws of Exponents:

1. x^y * x^z = x^(y+z) 2. (x^y)^z = x^(y*z) 3. (x*y)^z = x^z * y^z

where 0^0 = 1.

The trouble is, convenience aside, 0^0 = 0 also has these pleasing properties. It is also possible to formally construct or prove the existence of another binary function ^ on N such that

1. 0^0 = 0 2. x^0 = 1 for x=/=0 3. x^(y+1) = x^y * x

and to derive exactly the same Laws of Exponents as above. And, by Theorem 2, both functions are identical except for the value of 0^0.

In the absence of any logically compelling reasons (if mere convenience is not enough for the skeptics), 0^0 must be considered ambiguous (undefined or indeterminate). But, as I have formally proven here, there are infinitely many binary functions ^ on N that share the following characteristics:

1. x^0 = 1 for x=/=0 2. x^(y+1) = x^y * x

And that, from these properties, it is possible to derive only slight modified versions of the above Laws of Exponents:

(As one reader pointer out here, it is possible to tighten these stipulations so that other cases are covered, e.g. 0^2 * 0^3 = 0^(2+3).)

Clearly, it bothers some readers here that there is not just one, but infinitely many functions on N that fit all our requirements for exponentiation (i.e. for repeated multiplication). As I have shown here, however, this notion can be put on a rigorously sound mathematical footing using only the axioms and rules of ordinary logic and set theory. I, therefore, feel justified in defining exponentiation on N as a binary function ^ such that